leetcode Range Sum Query

题目连接

https://leetcode.com/problems/range-sum-query-mutable/  

Range Sum Query - Mutable

Description

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

The update(i, val) function modifies nums by updating the element at index i to val.

Example:

Given nums = [1, 3, 5]

sumRange(0, 2) -> 9
update(1, 2)
sumRange(0, 2) -> 8

Note:

1. The array is only modifiable by the update function.
2. You may assume the number of calls to update and sumRange function is distributed evenly.

线段树单点更新。。

class NumArray {
public:
	NumArray() = default;
	NumArray(vector<int> &nums) {
		n = (int)nums.size();
		if (!n) return;
		arr = new int[n << 2];
		built(1, 1, n, nums);
	}
	~NumArray() { delete []arr; }
	void update(int i, int val) {
		update(1, 1, n, i + 1, val);
	}
	int sumRange(int i, int j) {
		return sumRange(1, 1, n, i + 1, j + 1);
	}
private:
	int n, *arr;
	void built(int root, int l, int r, vector<int> &nums) {
		if (l == r) {
			arr[root] = nums[l - 1];
			return;
		}
		int mid = (l + r) >> 1;
		built(root << 1, l, mid, nums);
		built(root << 1 | 1, mid + 1, r, nums);
		arr[root] = arr[root << 1] + arr[root << 1 | 1];
	}
	void update(int root, int l, int r, int pos, int val) {
		if (pos > r || pos < l) return;
		if (pos <= l && pos >= r) {
			arr[root] = val;
			return;
		}
		int mid = (l + r) >> 1;
		update(root << 1, l, mid, pos, val);
		update(root << 1 | 1, mid + 1, r, pos, val);
		arr[root] = arr[root << 1] + arr[root << 1 | 1];
	}
	int sumRange(int root, int l, int r, int x, int y) {
		if (x > r || y < l) return 0;
		if (x <= l && y >= r) return arr[root];
		int mid = (l + r) >> 1, ret = 0;
		ret += sumRange(root << 1, l, mid, x, y);
		ret += sumRange(root << 1 | 1, mid + 1, r, x, y);
		return ret;
	}
};